Electromagnetic Induction Ques 5
5. A circular disc of radius $ 0.2$ meter is placed in a uniform magnetic field of induction $\frac{1}{p}(Wb / m^{2})$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is:
[2008]
(a) $0.02 $ $Wb$
(b) $0.06 $ $Wb$
(c) $0.08 $ $Wb$
(d) $0.01 $ $Wb$
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Answer:
Correct Answer: 5.(a)
Solution:
- (a) Here, $B=\frac{1}{p}(Wb / m^{2})$
$\theta=60^{\circ}$
Area normal to the plane of the disc
$=pr^{2} \cos 60^{\circ}=\frac{pr^{2}}{2}$
Flux $=B \times$ normal area
$=\frac{0.2 \times 0.2}{2}=0.02 $ $Wb$
BETA
